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The topic of odd perfect numbers likely needs no introduction.

Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.

If $n$ is odd and $\sigma(n)=2n$, then we call $n$ an odd perfect number. Euler proved that a hypothetical odd perfect number must necessarily have the form $n = p^k m^2$ where $p$ is the special prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

Descartes, Frenicle, and subsequently Sorli conjectured that $k=1$ always holds. Dris conjectured that the inequality $p^k < m$ is true in his M. Sc. thesis, and Brown (2016) eventually produced a proof for the weaker inequality $p < m$.

Now, recent evidence suggests that $p^k < m$ may in fact be false.

THE ARGUMENT

Let $n = p^k m^2$ be an odd perfect number with special prime $p$.

Since $p \equiv k \equiv 1 \pmod 4$ and $m$ is odd, then $m^2 - p^k \equiv 0 \pmod 4$. Moreover, $m^2 - p^k$ is not a square (Dris and San Diego (2020)).

This implies that we may write $$m^2 - p^k = 2^r t$$ where $2^r \neq t$, $r \geq 2$, and $\gcd(2,t)=1$.

It is trivial to prove that $m \neq 2^r$ and $m \neq t$, so that we consider the following cases:

$$\text{Case (1): } m > t > 2^r$$ $$\text{Case (2): } m > 2^r > t$$ $$\text{Case (3): } t > m > 2^r$$ $$\text{Case (4): } 2^r > m > t$$ $$\text{Case (5): } t > 2^r > m$$ $$\text{Case (6): } 2^r > t > m$$

We can rule out Case (5) and Case (6), and under Case (1) and Case (2), we can prove that the inequality $m < p^k$ holds.

So we are now left with Case (3) and Case (4):

Under both cases left under consideration, we have $$(m - 2^r)(m - t) < 0$$ $$m^2 + 2^r t < m(2^r + t)$$ $$m^2 + (m^2 - p^k) < m(2^r + t)$$ $$2m^2 < m(2^r + t) + p^k.$$

Since we want to prove $m < p^k$, assume to the contrary that $p^k < m$. We get $$2m^2 < m(2^r + t) + p^k < m(2^r + t) + m < m(2^r + t + 1)$$ which implies, since $m > 0$, that $$2m < 2^r + t + 1.$$

Here then is our question:

Will it be possible to derive a contradiction from the inequality $$2m < 2^r + t + 1,$$ under Case (3) and Case (4) above, considering that $2m$ is large? (In fact, it is known that $m > {10}^{375}$.)

Jose Arnaldo Bebita Dris
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  • So $m$ is abundant ? – Roddy MacPhee Jun 14 '21 at 01:22
  • Do you have a proof that $m$ must be abundant, from the constraints given in my question, @RoddyMacPhee? I would be interested to see it. – Jose Arnaldo Bebita Dris Jun 14 '21 at 02:52
  • Misread thought $m =2^rt$ for some reason then $2m<2^r+t+1$ would have implied it. – Roddy MacPhee Jun 14 '21 at 10:53
  • Assume to the contrary that $m = 2^r t$. Then $m^2 - p^k = 2^r t = m$, which is equivalent to $m(m - 1) = m^2 - m = p^k$. But this last equation contradicts $\gcd(p,m)=1$. Hence, we are sure that $m \neq 2^r t$, @RoddyMacPhee. – Jose Arnaldo Bebita Dris Jun 14 '21 at 12:13
  • Yeah still holds by multiplicative nature of sigma though... Just came up with wrong reasoning ... – Roddy MacPhee Jun 14 '21 at 13:20
  • Sorry, but I do not understand, @RoddyMacPhee. What equation / inequality follows from $\sigma(x)=\sigma_1(x)$ being multiplicative, where $x$ is a positive integer? – Jose Arnaldo Bebita Dris Jun 14 '21 at 23:45
  • If $p\not\mid m$ then $\sigma(n)=\sigma(p^k)\sigma(m)$ and since $\sigma(p^k)<2p^k$ we can at least say $sigma(m)>m$ – Roddy MacPhee Jun 14 '21 at 23:49
  • Thank you for your comment, @RoddyMacPhee. We are sure that $p \nmid m$, since $\gcd(p, m)=1$. Also, $n = p^k m^2$, so what you have is actually the equation $2p^k m^2=2n=\sigma(n)=\sigma(p^k)\sigma(m^2)$. Then we have the inequality $p^k < \sigma(p^k) < \frac{5p^k}{4}$, which finally gives $\frac{8m^2}{5}< \sigma(m^2) < 2m^2$. (In particular, we also know that $m < \sigma(m) < 2m$, but that follows from the fact that $m > 1$ and $m$ is deficient, being a proper factor of the perfect number $n = p^k m^2$.) – Jose Arnaldo Bebita Dris Jun 14 '21 at 23:58
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    $p^k\lt m$ is equivalent to $m\lt\dfrac{1+\sqrt{1+2^{r+2}t}}{2}$ which is better than $2m\lt 2^r+t+1$ since $\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\lt \dfrac{2^r+t+1}{2}$ is equivalent to $1\lt (2^r-t)^2$ which does hold. – mathlove Jun 18 '21 at 15:11
  • Thank you for your time and attention, @mathlove! I would be interested to see the details of your proof for the biconditional $$p^k < m \iff m < \frac{1 + \sqrt{1 + 2^{r+2} t}}{2}.$$ – Jose Arnaldo Bebita Dris Jun 18 '21 at 22:34
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    Since $ p^k = m^2-2^r t$, we have $p^k\lt m$$\iff m^2-2^r t\lt m$$\iff m^2-m-2^rt\lt 0$$\iff m\lt\dfrac{1+\sqrt{1+2^{r+2}t}}{2}$ – mathlove Jun 19 '21 at 05:06
  • Please do write out your last comment as an answer, @mathlove, so that I can upvote. Thanks! – Jose Arnaldo Bebita Dris Jun 20 '21 at 08:00
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    I've just converted my comment into an answer in which I added more thoughts. – mathlove Jun 20 '21 at 11:21

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On OP's request, I am converting my comment into an answer.

  • $p^k\lt m$ is equivalent to $$m\lt\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\tag7$$ since we have$$\begin{align}p^k\lt m&\iff m^2-2^rt\lt m \\\\&\iff m^2-m-2^rt\lt 0 \\\\&\iff m\lt\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\end{align}$$
  • $(7)$ is better than $2m\lt 2^r+t+1$ since $$\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\lt \frac{2^r+t+1}{2}\tag8$$ holds.

    To see that $(8)$ holds, note that $$\begin{align}(2)&\iff \sqrt{1+2^{r+2}t}\lt 2^r+t \\\\&\iff 1+2^{r+2}t\lt 2^{r+1}+2^{r+1}t+t^2 \\\\&\iff (2^r-t)^2\gt 1 \\\\&\iff |2^r-t|\gt 1\end{align}$$ which does hold.
  • We can say that $$\bigg(\dfrac{1+\sqrt{1+2^{r+2}t}}{2}-t\bigg)\bigg(\dfrac{1+\sqrt{1+2^{r+2}t}}{2}-2^r\bigg)\lt 0\tag9$$ since $$\begin{align}(9)&\iff \bigg(\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\bigg)^2-\dfrac{1+\sqrt{1+2^{r+2}t}}{2}(t+2^r)+2^rt\lt 0 \\\\&\iff \frac{1+\sqrt{1+2^{r+2}t}+2^{r+1}t}{2}-\dfrac{1+\sqrt{1+2^{r+2}t}}{2}(t+2^r)+2^rt\lt 0 \\\\&\iff 1+\sqrt{1+2^{r+2}t}+2^{r+1}t-(1+\sqrt{1+2^{r+2}t})(t+2^r)+2^{r+1}t\lt 0 \\\\&\iff 2^{r+2}t-2^r-t+1\lt (t+2^r-1)\sqrt{1+2^{r+2}t} \\\\&\iff (2^{r+2}t-2^r-t+1)^2\lt (t+2^r-1)^2(1+2^{r+2}t) \\\\&\iff 2^{r + 2} t (2^r - t - 1) (2^r - t + 1)\gt 0 \\\\&\iff (2^r-t)^2\gt 1 \\\\&\iff |2^r-t|\gt 1\end{align}$$ which does hold.
  • It follows from $(7)(9)$ that if $p^k\lt m$ with $(m-t)(m-2^r)\lt 0$, then $$\min(t,2^r)\lt m\lt\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\lt \max(t,2^r)$$
mathlove
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  • Is it obvious to you that $\left|2^r - t\right| \neq 1$? Would you mind sharing your proof for this assertion, @mathlove? The proof for $\left|2^r - t\right| \neq 1$ that I know is complicated, and considers Case (3) and Case (4) separately. It also uses ideas from this closely related MSE question, as well as methods from the preprint mentioned there. – Jose Arnaldo Bebita Dris Sep 30 '21 at 08:57
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    @Jose Arnaldo Bebita Dris : Suppose that $|2^r-t|=1$ which implies that $2^r,t$ are consecutive integers. Then, there is no integer $m$ such that $\max(2^r,t)\gt m\gt \min(2^r,t)$, a contradiction. – mathlove Sep 30 '21 at 09:31
  • That is indeed a very simple and at the same time elegant proof for $\left|2^r - t\right| \neq 1$, @mathlove! Thank you for sharing. =) – Jose Arnaldo Bebita Dris Sep 30 '21 at 10:20
  • Does your proof here for $p^k < m \implies |2^r - t| \neq 1$ effectively invalidate any further attempt at proving my conjectured implication $p^k < m \implies |2^r - t| = 1$, as detailed out in this MO question, @mathlove? – Jose Arnaldo Bebita Dris Dec 24 '21 at 00:52
  • @Jose Arnaldo Bebita Dris : I think yes since it is true that $p^k\lt m\implies (3)\ \text{or}\ (4)\implies |2^r-t|\not=1$. By the way, I think it is wrong to say that under (3) or (4), $|2^r-t|=1$ is sufficient for $p^k\lt m$ to hold since under (3) or (4), one has $|2^r-t|\not=1$ in the first place. – mathlove Dec 24 '21 at 04:08
  • I did obtain, under (3) or (4), that $p^k < m|2^r - t|$, which means that $|2^r - t| = 1$ is indeed sufficient for $p^k < m$ to hold, @mathlove. – Jose Arnaldo Bebita Dris Dec 24 '21 at 04:22
  • What I meant was that $|2^r - t| = 1 \implies |2^r - t| \neq 1$ just leads to the absolute value being not equal to $1$, and that it is not fundamentally wrong to assert that the equation where the absolute value equals $1$ is a sufficient condition for $p^k < m$, under (3) or (4), @mathlove. – Jose Arnaldo Bebita Dris Dec 24 '21 at 04:47
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    @Jose Arnaldo Bebita Dris : I should have said that it is vacuously true that under (3) or (4), $|2^r−t|=1$ is sufficient for $p^k<m$ to hold. – mathlove Dec 24 '21 at 04:55
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    By the way, I invite you to post your comment that $p^k < m \implies (3) \text{ or } (4) \implies |2^r - t| \neq 1$ as an actual answer to the MO question that I linked in my comment above. Thank you, @mathlove. – Jose Arnaldo Bebita Dris Dec 24 '21 at 04:56
  • No worries, @mathlove, I perfectly understood! – Jose Arnaldo Bebita Dris Dec 24 '21 at 04:58